To determine the geometry of a molecule, we need to find the steric number (number of bond pairs + lone pairs) around the central atom and apply VSEPR theory.

Step 1: Analyze the central atom and its valence electrons.

• A) ${\mathrm{IF}}_3$: Iodine (I) is the central atom. It has 7 valence electrons. It forms 3 single bonds with F atoms.
• B) ${\mathrm{PCl}}_3$: Phosphorus (P) is the central atom. It has 5 valence electrons. It forms 3 single bonds with Cl atoms.
• C) ${\mathrm{NH}}_3$: Nitrogen (N) is the central atom. It has 5 valence electrons. It forms 3 single bonds with H atoms.
• D) ${\mathrm{BF}}_3$: Boron (B) is the central atom. It has 3 valence electrons. It forms 3 single bonds with F atoms.

Step 2: Calculate the number of lone pairs and steric number.

• A) ${\mathrm{IF}}_3$: Valence electrons = 7. Electrons used in bonding = 3. Remaining electrons = 4. Number of lone pairs = $4/2=2$. Steric number = 3 (bond pairs) + 2 (lone pairs) = 5.
• B) ${\mathrm{PCl}}_3$: Valence electrons = 5. Electrons used in bonding = 3. Remaining electrons = 2. Number of lone pairs = $2/2=1$. Steric number = 3 (bond pairs) + 1 (lone pair) = 4.
• C) ${\mathrm{NH}}_3$: Valence electrons = 5. Electrons used in bonding = 3. Remaining electrons = 2. Number of lone pairs = $2/2=1$. Steric number = 3 (bond pairs) + 1 (lone pair) = 4.
• D) ${\mathrm{BF}}_3$: Valence electrons = 3. Electrons used in bonding = 3. Remaining electrons = 0. Number of lone pairs = $0/2=0$. Steric number = 3 (bond pairs) + 0 (lone pairs) = 3.

Step 3: Determine the molecular geometry based on VSEPR theory.

• A) ${\mathrm{IF}}_3$: Steric number 5 with 3 bond pairs and 2 lone pairs corresponds to T-shaped geometry.
• B) ${\mathrm{PCl}}_3$: Steric number 4 with 3 bond pairs and 1 lone pair corresponds to trigonal pyramidal geometry.
• C) ${\mathrm{NH}}_3$: Steric number 4 with 3 bond pairs and 1 lone pair corresponds to trigonal pyramidal geometry.
• D) ${\mathrm{BF}}_3$: Steric number 3 with 3 bond pairs and 0 lone pairs corresponds to trigonal planar geometry.

Therefore, ${\mathrm{BF}}_3$ has trigonal planar geometry.

Correct Answer: (D)